Final answer:
To solve the differential equation y'' + 4y = g(t) with initial conditions y(0) = 1 and y'(0) = 0, we can divide the problem into two cases based on the given forcing function. For the interval 0 ≤ t < π, the solution is y(t) = c1*cos(2t) + c2*sin(2t) + t/2 + 1. For the interval π ≤ t ≤ 2π, the solution is y(t) = c3*cos(2t) + c4*sin(2t).
Step-by-step explanation:
To solve the differential equation y'' + 4y = g(t) with initial conditions y(0) = 1 and y'(0) = 0, we need to solve for y as a function of t. Since g(t) is piecewise-defined, we'll solve for y in each interval separately.
For the interval 0 ≤ t < π, g(t) = 4. In this case, the differential equation becomes y'' + 4y = 4. The general solution to this homogeneous differential equation is y(t) = c1*cos(2t) + c2*sin(2t). To find the particular solution, we can use the method of undetermined coefficients and assume y(t) = At + B. Substituting this into the differential equation gives A = 0 and B = 2/4 = 1/2. Therefore, the solution for this interval is y(t) = c1*cos(2t) + c2*sin(2t) + t/2 + 1.
For the interval π ≤ t ≤ 2π, g(t) = 0. In this case, the differential equation becomes y'' + 4y = 0. The general solution to this homogeneous differential equation is y(t) = c3*cos(2t) + c4*sin(2t). Since g(t) = 0, there is no need to find a particular solution. Therefore, the solution for this interval is y(t) = c3*cos(2t) + c4*sin(2t).